Enthalpy

ENTHALPY
(Internal energy + energy from pressure & volume)

The area of focus for this tutorial is Force & Energy and Mathematics. Actually, this picture is good to illustrate the "First Law of Thermodynamics," which says that energy cannot be created or destroyed. The sun doesn't create energy; it's just converting mass from hydrogen atoms into light energy. That energy hitting the Earth becomes thermal energy (warmth) and kinetic energy (movement like wind, rivers, ocean currents, etc). Regarding the pyramid, the energy from the food calories that workers ate are still stored as potential energy in the heavy stones that sit high on the pyramid. So energy is not created or destroyed; it just changes form. We can take advantage of that fact when figuring out how much energy is released or absorbed from chemical reactions.

Again, the energy in the universe is constant, but when you zoom in on one little spot, you can see its energy levels going up or down. If the energy of that spot has gone down, then that spot must have transferred energy away from itself in the form of heat or as a force that pushed something across a distance. In other words, that drop of energy must have gone into its surroundings, which is the rest of the universe.

Let's zoom in on a part of the universe and focus on the black powder wrapped in white cloth that sits behind the black cannon ball. The area of focus (black powder in this case) is called the "system" and everything else is called the surroundings. Right now the black powder has a lot of chemical energy in it, which is a form of potential energy where the + and - charges (protons and electrons) are separated farther from each other than is possible after rearrangement of the atoms. The black powder also has some thermal energy because the molecules in the gunpowder are moving (vibrating). Together, we call these energies its "Internal Energy". That's represented by the letter "E".

When a cannon fires, we see the internal energy of the black powder dropping drastically. Originally there was chemical energy in the bonds (attraction of electrons to protons) of the ingredients in the gunpowder (charcoal, sulfur, and potassium nitrate). Now we see energy in the form of pressure and volume of expanding gases. The boom indicates energy in the air also in the form of pressure. There's also a lot of heat energy being transferred to the barrel of the cannon and to the air. The light from the muzzle flash is another form of energy. Of course, there's the kinetic energy of the heavy cannon ball flying away, which was suppose to be the only energy desired. Again, we started with a lot of chemical energy in the black powder plus some thermal energy (Together that's the Internal Energy), but most of that energy has gone into the surroundings.

Afterwards, the combustion products of the black powder have much less chemical energy, and after cooling down to ambient air temperature, the thermal energy is the same as before. So the powder's Internal Energy dropped drastically due to loss of chemical energy. The change in chemical energy was lost as heat and used up in providing a force (a pressure) that pushed the air and cannon ball out of the barrel. Note: When a force is applied over a distance that is called Work Energy, represented by a "w". So part of the energy of the powder provided the work energy needed to build up pressure and expand gases out the barrel. Mathematically, the change in its internal energy is represented by E_final-E_intial or written as ΔE, where the Δ means the change or difference between initial and final state. The heat transferred (lost) is represented by the letter "q". The work energy is represented by "w". So the formula for change of internal energy is
ΔE = q + w
In other words if you add up the heat transfer and the work done, that will account for the change in the internal energy of a system. Note: If the system performs work on the surroundings (in this case it is compressed air and shot a cannon ball) the work is expressed with a negative number.

Since work is a force times distance, let's choose 100 pounds of force times the distance of 5 inches. Notice that if we had a gas pushing against 50 pounds per square inch and that is multiplied by a volume change of 10 cubic inches, we end up with the same answer of 500 pound·inches of energy.

	A	B	C	D	E	F	G
1	10	pounds	5	inches	=	500	pound·inches

2	Pressure to overcome		Change in Volume			Energy converted
3	5	pounds	10	in³=in x in x in	=	500	pound·inches
4		in²=in x in

For many reactions, it is easier to measure pressure and volume, so Work (w) is commonly replaced with PV (pressure times volume). The units are the same. So the formula is often written as :
ΔE = q - Δ(PV)
A minus sign is used if the system did work on the surroundings (pushed on it across a distance).

In the other earlier tutorial on energy I talked about the heat energy stored in materials such as gold and water as shown. Specific Heat was also discussed and we talked about the calories or joules that one gram of a substance would absorb or release when its temperature rose or fell. However, we didn't consider gases in that tutorial. Gases have additional energy because of their pressure and volume. Remember the ideal gas law, PV=nRT? It was never mentioned, but pressure times volume (PV) calculates energy. So PV=nRT is saying the energy from the pressure and volume of the gas is equal to the moles of that gas times its temperature times the R constant.

A term that encompasses both of these energies is called "Enthalpy". "thalpy" comes from the Greek word, "thalpein" meaning "to heat" and the "En" means "put into". So together they refer to putting heat into something, which is the "heat energy or heat capacity" aspect of enthalpy.

Remember this air-powered bicycle. We did calculations about the mass of gas in the tanks, but not its energy. So enthalpy takes in consideration its energy due to its pressure and volume plus its heat energy (heat capacity). The formula is:
H = E + PV
H represents enthalpy. E represents "Internal Energy" which is related to heat capacity (or heat energy). PV is pressure times volume. Note, that in some books you see "U" used to represent "Internal Energy". Internal energy is both the kinetic energy of the movement of the atoms or molecules plus the potential energy of + and - charges that are separated. These factors also influence heat capacity (joules per °C) and specific heat (joules per °C per gram). Internal Energy is just joules.

Most chemistry is done under the constant pressure of the atmosphere. When there's a constant pressure during a reaction, the enthalpy equation is simplified.

Besides just the absolute enthalpy energy, chemistry is more concerned about reactions where energy is released or absorbed. So when there's a change, our enthalpy equation becomes:
ΔH = ΔE + Δ(PV)
Above we said the change in internal energy is due to transfer of heat energy (q) plus the work energy done by the system on its surroundings as being:
ΔE = q - Δ(PV)
When you substitute ΔE in the enthalpy equation with the q-Δ(PV) you get
ΔH = q - Δ(PV) + Δ(PV)
What this means is that the - Δ(PV) + Δ(PV) will cancel out if either the pressure or volume stays constant. In many situations where chemical reactions take place, this is true. This means that the change in enthalpy is equal to the amount of heat energy transferred (q). So the equation is simplified since the change of enthalpy is equal to the energy transferred in or out of the system.
ΔH = q
ΔH = H_final - H_initial
ΔH = H_products - H_reactants

q = H_products - H_reactants

What this means is that when we measure the heat released or absorbed by a reaction, we are getting insight into the energy of the products and the reactants.

If heat is released during the reaction then:
... the reaction is exothermic.
... q is negative.
... the reactants have more enthalpy energy than the products.
... the products generally will have stronger bonds than the bonds in the reactants.
... the products are more stable than the reactants.
... the electrons in the products are on average closer to the protons than the electrons in the reactants.

If heat is absorbed during the reaction then:
... the reaction is endothermic.
... q is positive.
... the reactants have less enthalpy energy than the products.
... the products generally will have weaker bonds than the bonds in the reactants.
... the products are less stable than the reactants.
... the electrons in the products are on average farther from the protons than the electrons in the reactants.

Sample Problem:
Earlier we showed the equation for the change of internal energy:
ΔE = q + w
q is heat transferred and "w" is work done, which is a force times distance. The device on the left is a "bomb calorimeter" which burns substances inside a rigid steel container submerged in water. Burning normally creates gases that will expand; however, the rigid container prevents any expansion, so there's no work done (no force x distance) on the surroundings by the burning gases. Therefore, "w" is zero and the equation simplifies to ΔE = q

The way these bomb calorimeters work is that they are calibrated such that it is known how many joules (or calories) on energy is needed to raise them 1 degree Celsius. When a substance of known mass is placed in a metal chamber in a bomb calorimeter, it is surrounded by pure oxygen and then ignited using an electric filament. The heat from the combustion heats up the bomb calorimeter and a stirrer makes sure the heat is distributed evenly. If a bomb calorimeter is rated at 12.1 joules per °C (or degree Kelvin, which is the same increment), then all you have to do is multiply the temperature rise by 12.1 joules/°C to get the energy in the substance. Divide that energy by the gram mass of the substance and you get the joules per gram for the combustion reaction of that substance.

Hess's Law

Hess's Law is named after Germain Hess, a Russian doctor and chemist. This law is used to calculate energy changes in reactions that are not easy to do directly. It depends on the fact that the energy difference between the initial reactants and the final products does not depend on how the reactants turned into the products, just that eventually they did. It is similar to the altitude difference between the top of a mountain and the lake below. It doesn't matter if you go from the lake straight up to the top or climbed to another mountain first. The altitude difference from the top of the mountain to the lake's surface is the same.

For example, let's say we need to know how much heat will be produced when carbon is burned to form carbon monoxide.
2C_(s) + O₂_(g) --> 2CO_(g)

That is hard to do because one needs to burn carbon in limited oxygen, but that will still form a mixture of carbon dioxide and carbon monoxide. However, we can figure this using two other reactions that were easy to measure. The first reaction is from burning carbon in an excess of oxygen:
C_(s) + O₂_(g) --> CO₂_(g)

The change of enthalpy, ΔH, for this reaction is -393.7 kilojoules per mole of carbon.

Remember -393.7 kilojoules per mole of carbon is just words and symbols. I like to visualize what this means. A mole of carbon is 12 grams, the weight of 2 1/2 nickels. That's probably about half a tablespoon of carbon. Now that I have the substance visualized, I move on to the energy. The "-" means energy is given off, so I see this reaction like a fire which creates heat. Whenever I see "joules" I think of watts because 1 joule per second is 1 watt. So the 393.7 kilojoules (393,700 joules) is the energy of about 400,000 watts run for one second. A home air conditioner uses about 4,000 watts. So that means this 1/2 tablespoon of carbon can generate the energy to run 100 air conditioners for one second or 1 air conditioner for 100 seconds. I also like to turn joules into food Calories. A rough conversion is to divide joules by 4 to get calories. So this is about 100 kilocalories or 100 food calories per mole or per 12 grams. That's in between the calories for sugar (5 calories per gram) and fats (9 calories per gram). So burning is about the same calories as icing. However, we can't digest pure carbon, so to us there's no calories, but burning does produce calories.

A second reaction is the burning of carbon monoxide:
2CO_(g) + O₂_(g) --> 2CO₂_(g)

The change of enthalpy, ΔH, is -283.3 kilojoules per mole of CO.
Again the visualization is a mole of a CO gas (any gas actually) is about 22 liters . That's about 5 gallons of this poisonous gas. The negative sign means it produces energy when burned. The 283.3 kilojoules is about 2/3 of what we visualized for the carbon combustion above.

The next task is to arrange these two reactions and add them so that they create the reaction we are looking for. Row 8 is the final target reaction. Rows 2 and 5 are the given reactions above.

	A	B	C	D	E	F	G	H	I
1		The below given reaction only has 1 carbon. We need it doubled to match target reaction on row 8.					multiplier
2	Given	C_(s) + O_2(g)	-->	CO_2(g)	-393.7	kilojoules/mole
3	Row 2 Multiplied	2C_(s) + 2O_2(g)	-->	2CO_2(g)	-393.7	kilojoules/mole	2	-787.4	kJ/mol
4		The below reaction has CO on the left. We need it reversed so that CO is on the right like the target reaction
5	Given	2CO_(g) + O₂_(g)	-->	2CO_2(g)	-283.3	kilojoules/mole	"-" goes to + when reversed
6	Row 4 Reversed	2CO₂_(g)	-->	2CO_(g) + O_2(g)	+283.3	kilojoules/mole	1	+283.3	kJ
7	Below is the target reaction that we need to find the change in enthalpy (ΔH). Only Rows 3 and 6 are added. 2CO₂ in B6 cancels the 2CO₂ in D3. O₂ in D6 cancels one of the O₂ in B3. So after canceling, the 2 reactions add up to Row 8. The ΔH's add up to -504.1 kJ.
8	Target reaction	2C_(s) + O_2(g)	-->	2CO_(g)				-504.1	kJ

The above given reactions were multiplied or reversed in order to achieve the target reaction when added. The ΔH of the adjusted reactions were added to find the final ΔH of the target reaction. So we showed that even though carbon was turned into carbon dioxide first (Row 3) and then the carbon dioxide was decomposed to carbon monoxide, it's the same energy-wise as simply going straight from carbon to carbon monoxide.

The next section will take this idea of creating virtual reactions from standard reactions. It involves some extra symbols, so remember my warning at the beginning the class about being a symbologist like the characters in these clue-finding mystery movies. The next symbol to learn is ΔH°_f . The "f" actually goes below the degree symbol but that's not easy to do on a Web page. No, the ° symbol does not mean temperature degrees. It means the reaction took place 1 atmosphere of pressure and at 25°C (77°F). The "f" means the enthalpy of formation. In other words the enthalpy of forming the products from the reactants. So it's the same as we've been using. ΔH°_f has the units of kilojoules per mole (kJ/mol).

There are tables that show the enthalpy of formation for many compounds. You can uses these to figure out if a reaction that involves these compounds is going to be exothermic or endothermic and by how much. So it tells you if a reaction is likely to occur on its own or will it require a lot of energy. All of this effort of learning these concepts is that you can design something on paper and when you build it, it has a good chance of actually working. A lot of inventors who fail, just didn't take the time to figure if the reactions or energies involved would actually make it work.

The example used in the textbook is the burning of methane (main ingredient of natural gas).

CH₄_(g) + 2O₂_(g) --> CO₂_(g) + 2H₂O_(l)

Each of these are considered separately, and the elements that make them are used to form them. For example, methane is made from carbon and hydrogen using this equation.
C_(s) + 2H₂_(g) --> CH₄_(g) -75kJ/mole

The table of enthalpies of formation list this as -75kJ/mole of methane. In our desired reaction above, the methane is on the left, so we need to reverse this standard reaction and change the -75 to +75.
CH₄_(g) -->C_(s) + 2H₂_(g) +75kJ/mole

Below we can do all of them in spreadsheet table format.

	A	B	C	D	E	F	G	H	I
1	Target reaction where we want to know what the change in enthalpy is.	CH_4(g) + 2O₂_(g)	-->	CO_2(g) + H₂O_(l)
2		The below reaction is the standard formation of CH₄. In our target reaction above, CH₄ is on the left, so we need to reverse the below reaction					X
3	CH₄ Enthalpy of Formation reaction	C_(s) + 2H_2(g)	-->	CH_4(g)	-75	kilojoules/mole
4	Row 2 Reversed	CH_4(g)	-->	C_(s) + 2H_2(g)	+75	kilojoules/mole	1	+75	kJ/mol
5	Oxygen in standard state has zero Enthalpy of Formation	O₂_(g)	-->	O_2(g)	0	kilojoules/mole	1	0	kJ/mol
6	CO₂ Enthalpy of Formation reaction	C_(s) + O_2(g)	-->	CO_2(g)	-394	kilojoules/mole	1	-394	kJ/mol
7	The problem is that our target reaction has two H₂O molecules. So we double this reaction	H_2(g)+ ½O_2(g)	-->	H₂O_(l)	-286	kilojoules/mole
8	This is doubled to match target reaction	2H_2(g)+ O_2(g)	-->	2H₂O_(l)	-286	kilojoules/mole	2	-572	kJ/mol
9	In reactions in yellow, items on both sides cancel leaving our target reaction	CH_4(g) + 2O₂_(g)	-->	CO_2(g) + H₂O_(l)		ΔH's added		-891	kJ/mol
10	Items in red are on both sides, so they cancel each other. The remaining reactants and products add up to make the target reaction. The ΔH's add up to -891 kJ/mol which is a very exothermic reaction, which we expect since methane is a common fuel source.

Well you survived a tutorial on enthalpy. In the Chapter Textbook problems, there will be more problems to reinforce what you learned above.